Q-I-J-are-two-ideals-of-commutative-ring-R-prove-that-I-J-I-J-m-n-note-I-x-R-n-N-x-n-I-

Question Number 174684 by mnjuly1970 last updated on 08/Aug/22

Q : I, J a r e t w o i d e a l s o f c o m m u t a t i v e

r i n g, (R, \oplus,) . p r o v e t h a t :

\sqrt{I \cap J} \overset{?}{=} \sqrt{I} \cap \sqrt{J} m . n

n o t e : \sqrt{I} = {x \in R ∣ \exists n \in N, x^{n} \in I}

Answered by mindispower last updated on 09/Aug/22

I \cap J i d e a l

\sqrt{I \cap J} = {x \in R ∣ \exists n \in N x^{n} \in I \cap J}

x^{n} \in I \Rightarrow x \in \sqrt{I}

x^{n} \in J \Rightarrow x \in \sqrt{j}

\Rightarrow x \in \sqrt{I} \cap \sqrt{J} \Rightarrow \sqrt{I \cap J} \subset \sqrt{I} \cap \sqrt{J}

x \in \sqrt{I} \cap \sqrt{J} \Rightarrow \exists n, m ∣ x^{n} \in I, x^{m} \in J

x^{n + m} \in I “ x^{n} . x^{m} \in x^{m} . I = I, “ x^{n + m} \in J, x^{m + n} = x^{n} . x^{m} \in x^{n} J = J}

\Rightarrow x^{n + m} \in I \cap J \Rightarrow x \in \sqrt{I \cap J} \Rightarrow \sqrt{I} \cap \sqrt{J} \subset \sqrt{I \cap J}

\sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}

Commented by mnjuly1970 last updated on 14/Aug/22

t h a n k s a l o t s i r p o w e r \dots